$\int_0^{2\pi} e^{\cos(\phi)}\cos(\phi - \sin(\phi)) d\phi$ via contour integration

Question

Can anyone help me calculating this integral using contour integration?

$\int_0^{2\pi} e^{\cos(\phi)}\cos(\phi - \sin(\phi)) d\phi$

I've used the subctraction formula of the cosine:

$$\cos(\phi - \sin(\phi)) = \cos(\phi)\cos(\sin(\phi)) + \sin(\phi)\sin(\sin(\phi))$$

then substituting $z = e^{i\phi}$:

$\displaystyle \int_0^{2\pi} e^{\cos(\phi)}\cos\left(\phi - \sin(\phi)\right) d\phi$ $\displaystyle =\int_0^{2\pi} e^{\frac{1}{2} \left(z + \frac{1}{z}\right)}(\frac{1}{2} \left(z + \frac{1}{z}\right)\cos\left(\frac{1}{2} \left(z - \frac{1}{z}\right)\right) + \frac{1}{2i} \left(z - \frac{1}{z}\right)\sin\left(\frac{1}{2i} \left(z - \frac{1}{z}\right)\right)) d\phi$

I've used again trigonometric formulas to break the cos and sin and then I've substitute the exponential, the sine and the cosine with their Taylor and Laurent series in $z = 0$ (which seems to me the only singularity), and tried to find the residue isolating the $1/z$ coefficient. Is that a correct way? No matter how hard I try I can't get the result $2 \pi$.

Answer 1

Your integral is the real part of

$$\int_0^{2\pi} e^{\cos \phi}e^{i(\phi - \sin \phi)}\, d\phi = \int_0^{2\pi} e^{\cos \phi - i\sin \phi}e^{i\phi}\, d\phi = \int_0^{2\pi} e^{e^{-i\phi}}e^{i\phi}\, d\phi$$

The rightmost integral can be represented as

$$\oint_{|z| = 1} e^{z^{-1}} z\frac{dz}{iz} = \frac{1}{i}\oint_{|z| = 1} e^{z^{-1}}\, dz.$$

By the residue theorem,

$$\oint_{|z| = 1} e^{z^{-1}}\, dz = 2\pi i \underset{z = 0}{\operatorname{Res}} e^{z^{-1}} = 2\pi i.$$

Hence

$$\frac{1}{i}\oint_{|z| = 1} e^{z^{-1}}\, dz = 2\pi,$$

and consequently your integral evaluates to $2\pi$.

Answer 2

First note that $\cos\alpha=\cos(-\alpha)$, so your integral can be written $$\eqalign{I &=\int_0^{2\pi}e^{\cos\phi}\cos(\sin\phi-\phi)\,d\phi\cr &={\rm Re}\int_0^{2\pi}e^{\cos\phi+i\sin\phi-i\phi}\,d\phi\cr &={\rm Re}\int_0^{2\pi}e^{e^{i\phi}}e^{-i\phi}\,d\phi\ .\cr}$$ Now substitute $z=e^{i\phi}$ to get an integral around the unit circle, $$\eqalign{I &={\rm Re}\int_C \frac{e^z}{z}\,\frac{dz}{iz}\cr &=2\pi\,{\rm Res}\Bigl(\frac{e^z}{z^2},z=0\Bigr)\cr &=2\pi\ .\cr}$$